If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
The 8th term from the end (of 30 terms) = the (30 − 8 + 1) = 23rd term from the beginning.
Given: $a_{30} = 119$ and $a_{23} = 91$
Using $a_n = a + (n-1)d$:
$$a_{30} - a_{23} = 7d \implies 119 - 91 = 7d \implies 28 = 7d \implies \boxed{d = 4}$$
Finding first term $a$:
$$119 = a + 29 \times 4 \implies a = 119 - 116 = 3$$
Sum of all 30 terms:
$$S_{30} = \frac{n}{2}(a + l) = \frac{30}{2}(3 + 119) = 15 \times 122 = \boxed{1830}$$
Source: Chapter 5, Sections 5.2 and 5.3
---
Explanation
- Key step: The 8th term from the end in a 30-term AP is the $(30 - 8 + 1) = 23$rd term from the start. Students often miss this conversion — it's where most marks are lost.
- Use the difference of two $a_n$ expressions to eliminate $a$ and solve for $d$ directly.
- For sum, use $S = \frac{n}{2}(a + l)$ since both first and last terms are known — it's quicker than the other formula.
- Show each step clearly; all 3 marks are awarded for: finding the 23rd term position (½), finding $d$ (1), finding $S_{30}$ (1).